My lecturer has a passion for logs, and I'm reviewing some of her past papers and I found this question and I'm having a quite a difficult time dealing with, any help would be appreciated
$$ \sum_{n=3}^\infty \frac {(-1)^{n+1}}{n\log^p n \log \log n }.$$
Since I have to decide for what values of p the series behaves in a certain way my first instinct is to use the condensation test. Which gets me to;
$$ a_n = \frac {(-1)^{n+1}}{n\log^p n \log \log n } 2^n a_{2^n} = \frac {(-1)^{2n+1}}{n^2 \log^p2\log\log2}.$$
Is this the correct path to be on?
Outline: Let $p\gt 1$. After a while, $\log\log n\gt 1$, so after a while we have $$0\lt \frac{1}{n\log^p n\log\log n}\lt \frac{1}{n\log^p n}.$$ Recall that if $p\gt 1$, the series $\sum_3^\infty \frac{1}{n\log^p n}$ converges (Integral Test, or Cauchy Condensation). So by Comparison, our original series converges absolutely if $p\gt 1$.
For $p\le 1$, show in some way that the series does not converge absolutely when $p=1$, and use Comparison to conclude we do not have absolute convergence for any $p\le 1$.
As to conditional convergence, we do have it for any $p\le 1$, including negative values. Recall that for convergence. divergence, only what happens "in the long run" matters. We can if we wish disregard the first $10000$ terms. The Leibniz (Alternating Series) Test will work. It is easier for $0\le p\le 1$ than for negative $p$.
Remark: For Cauchy Condensation, note that it is a test for series with positive terms. The terms must also (for large enough $n$) decreasing in absolute value.
To use it on your original series, we replace $n$ by $2^m$ and multiply by $2^m$. We do the dtails, since they do not appear to be right in the OP.
The denominator is then $2^m \log^p(2^m)\log\log(2^m)$. The $\log^p(2^m)$ part is $(\log 2)^pm^p$. The $\log\log(2^m)$ part is of no real importance, but for the rcord it is $\log\log2+\log m$.
So when we take the reciprocal and multiply by $2^m$ we get $\dfrac{1}{(\log 2)^p m^p (\log\log 2+\log m)}$.